Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $p \neq 0$. $a = \dfrac{-p + 6}{p^2 + p - 42} \times \dfrac{p^2 + 10p + 16}{3p + 24} $
First factor out any common factors. $a = \dfrac{-(p - 6)}{p^2 + p - 42} \times \dfrac{p^2 + 10p + 16}{3(p + 8)} $ Then factor the quadratic expressions. $a = \dfrac {-(p - 6)} {(p - 6)(p + 7)} \times \dfrac {(p + 8)(p + 2)} {3(p + 8)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac {-(p - 6) \times (p + 8)(p + 2) } { (p - 6)(p + 7) \times 3(p + 8)} $ $a = \dfrac {-(p + 8)(p + 2)(p - 6)} {3(p - 6)(p + 7)(p + 8)} $ Notice that $(p - 6)$ and $(p + 8)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac {-(p + 8)(p + 2)\cancel{(p - 6)}} {3\cancel{(p - 6)}(p + 7)(p + 8)} $ We are dividing by $p - 6$ , so $p - 6 \neq 0$ Therefore, $p \neq 6$ $a = \dfrac {-\cancel{(p + 8)}(p + 2)\cancel{(p - 6)}} {3\cancel{(p - 6)}(p + 7)\cancel{(p + 8)}} $ We are dividing by $p + 8$ , so $p + 8 \neq 0$ Therefore, $p \neq -8$ $a = \dfrac {-(p + 2)} {3(p + 7)} $ $ a = \dfrac{-(p + 2)}{3(p + 7)}; p \neq 6; p \neq -8 $